Finite coverings by closed subschemes

Question

Let $X$ be a scheme. Assume we have two closed subschemes $Y_1$, $Y_2$ that cover $X$ set-theoretically. Are there closed subschemes $Y'_1$, $Y'_2$ with the same underlying sets, such that the natural map $Y'_1\amalg Y'_2\to X$ is schematically dominant? (Equivalently, we want the intersection of the ideal sheaves of $Y'_1$, $Y'_2$ to be zero).

This is true if $X$ is locally noetherian, as follows easily from the ''irredundant decomposition'' of EGA IV, (3,2.6) (a sheafification of primary decomposition).

But I cannot think of any other proof. Does anyone have a (non-noetherian) counterexample?

In the affine case, the problem translates as follows: Let $R$ be a ring with two ideals $I_1$, $I_2$ such that $I_1\cap I_2$ is contained in the radical. Find ideals $I'_j$ ($j=1,2$)such that $\sqrt{I'_j}=\sqrt{I_j}$ and $I'_1\cap I'_2=\{0\}$.

Answer 1

Here's a counterexample.

$R=k[x,y_1,y_2,...]/((xy_i)^{1+i}: i\ge1)$.

$I_1=(x), I_2=(y_1,y_2,...)$.

$I_1,I_2,I_1 \cap I_2$ are radical ideals and each element of $I_1 \cap I_2$ is nilpotent.

Some power of $x$, say $x^n$, is in $I'_1$. Some power of $y_n$, say $y_n^m$, is in $I'_2$. Then $x^ny_n^m \ne 0$ is in $I'_1 \cap I'_2$.